Let $M$ be a $2\times 2$ Hermitian matrix such that $M^{2} = I_{2}$ (the $2\times 2$ identity matrix) and define:
$$S = \frac{1}{2}\begin{pmatrix}
M & 0 \\
0 & M
\end{pmatrix}
$$
a $4 \times 4$ matrix. I want to prove that $S$ has eigenvalues $\{\pm \frac{1}{2}\}$ and $\operatorname{dim}\operatorname{ker}(S \pm \frac{1}{2}) = 2$. The argument goes as follows. Since $M^{2} = I_{2}$ and $M \neq I_{2}$, it follows that the eigenvalues of $M$ are $\{\pm 1\}$, with multiplicity of each eigenvalue being one. By this fact, the definition of $S$ and the invariance of spectrum, one obtains the results.
Now, I want to better understand the argument.
(1) The first part of the argument proves that the eigenvalues of $M$ are $\{\pm 1\}$. Since $M^{2} = I_{2}$ is Hermitian, then $M^{\dagger} = M = M^{-1}$, so $M$ is unitary. As a consequence, the eigenvalues of $M$ have absolute value $|\lambda| = \pm 1$. Moreover, since $M$ is Hermitian, it must have only real-valued eigenvalues, and this implies $\lambda = \pm 1$. I believe this is the reasoning behind the first part of the argument. However, why do we need $M \neq I_{2}$? And also, why can we conclude that the multiplicity of each eigenvalue is one?
(b) I don't quite understand the second part of the argument. How should I use the unitary invariance of the spectrum to obtain the eigenvalues of $S$? And why is $\operatorname{dim}\operatorname{ker}(S \pm \frac{1}{2}) = 2$?
Best Answer
Question/remark: "Now, I want to better understand the argument."
Response: Assume $Mv=\lambda v$ for some scalar $\lambda$ and $v \neq 0$. It follows
$$v=Iv=M^2v=M(Mv)=M(\lambda v)=\lambda(Mv)=\lambda^2v$$
hence $\lambda^2=1$ and $\lambda =1,-1$. You get two non-zero eigenvectors $u,v$ with
$$Mu=u, Mv=-v$$
and 4 eigen vectors for $S$: $v_1:=(u,0),v_2:=(v,0),v_3:=(0,u),v_4:=(0,v)$ with eigen value $\pm \frac{1}{2}$.The matrix $S\pm\frac{1}{2}$ has kernel $v_1,v_3$ or $v_2,v_4$, hence $ker(S\pm\frac{1}{2})$ has dimension $2$.